A particle executes SHM on a straight line path. The amplitude of oscillation is 2 cm. When the displacement of the particle from the mean position is 1 cm, the magnitude of its acceleration is equal to that of its velocity. Find the time period, maximum velocity and maximum acceleration of SHM.

Explanation

Here, `A= 2cm, y= 1cm, omega =?`
`V_("max") = ?, A_("max") =?`
At displacement y, `a= v` (value)
`therefore omega^(2)y = omega sqrt(A^(2) -y^(2))`
`therefore omega^(2)y = omega sqrt(A^(2)-y^(2))`
In `omega^(2)y^(2) = A^(2) -y^(2)`
Putting y=1 cm and A= 2 cm,
`omega^(2)(1)^(2) = (2)^(2) -(1)^(2)`
`omega^(2) = 4-1 implies omega = sqrt(3) rad s^(-1)`.
Periodic time `T= (2pi)/(omega) = (2pi)/(sqrt(3))s`
And maximum velocity,
`v_("max") = A omega`
`= 2 xx sqrt(3) cms^(-1)`
And maximum acceleration,
`a_("max") = A omega^(2)`
`= 2xx (sqrt(3))^(2) = 6cms^(-2)`.